AP EAMCET · PHYSICS · Electrostatics
An infinitely long thin straight wire has uniform linear charge density of \(\frac{1}{3} \mathrm{Cm}^{-1}\). Then, the magnitude of the electric intensity at a point \(18 \mathrm{~cm}\) away is (given \(\varepsilon_0=8.8 \times 10^{-12} \mathrm{C}^2 \mathrm{Nm}^{-2}\) )
- A \(0.33 \times 10^{11} \mathrm{NC}^{-1}\)
- B \(3 \times 10^{11} \mathrm{NC}^{-1}\)
- C \(0.66 \times 10^{11} \mathrm{NC}^{-1}\)
- D \(1.32 \times 10^{11} \mathrm{NC}^{-1}\)
Answer & Solution
Correct Answer
(C) \(0.66 \times 10^{11} \mathrm{NC}^{-1}\)
Step-by-step Solution
Detailed explanation
Charge density of long wire \(\begin{aligned} \lambda & =\frac{1}{3} \mathrm{C}-\mathrm{m} \\ \text { and } \quad r & =18 \times 10^{-2} \mathrm{~m} \end{aligned}\) From Gauss theorem…
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